Problem: Evaluate the triple integral. $ \int_{-1}^1 \int_0^2 \int_0^{z^2} x + 2y - 2z \, dx \, dz \, dy =$
Solution: We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_{-1}^1 \int_0^2 \int_0^{z^2} x + 2y - 2z \, dx \, dz \, dy \\ \\ &= \int_{-1}^1 \int_0^2 \left[ \dfrac{x^2}{2} + 2yx - 2zx \right]_0^{z^2} dz \, dy \\ \\ &= \int_{-1}^1 \int_0^2 \dfrac{z^4}{2} + 2yz^2 - 2z^3 \, dz \, dy \end{aligned}$ The second layer: $\begin{aligned} &\int_{-1}^1 \int_0^2 \dfrac{z^4}{2} + 2yz^2 - 2z^3 \, dz \, dy \\ \\ &= \int_{-1}^1 \left[ \dfrac{z^5}{10} + \dfrac{2yz^3}{3} - \dfrac{z^4}{2} \right]_0^2 dy \\ \\ &= \int_{-1}^1 \dfrac{16}{5} + \dfrac{16y}{3} - 8 \, dy \\ \\ &= \int_{-1}^1 \dfrac{16y}{3} - \dfrac{24}{5} \, dy \end{aligned}$ The third layer: $\begin{aligned} &\int_{-1}^1 \dfrac{16y}{3} - \dfrac{24}{5} \, dy \\ \\ &= \left[ \dfrac{8y^2}{3} - \dfrac{24y}{5} \right]_{-1}^1 \\ \\ &= \left(\dfrac{8}{3} - \dfrac{24}{5} \right) - \left( \dfrac{8}{3} + \dfrac{24}{5} \right) \\ \\ &= \dfrac{-48}{5} \end{aligned}$ In conclusion: $ \int_{-1}^1 \int_0^2 \int_0^{z^2} x + 2y - 2z \, dx \, dz \, dy = \dfrac{-48}{5}$